Mathematics | Studies, essays, thesises » Security Analysis Of High Building Parachute Jumping

Source: http://www.doksinet Security Analysis Of High Building Parachute Jumping 王晓骎 谢海润 谢浩然 许 瑞 I. Abstract: Recently, a disastrous fire attacked a teachers’ apartment on Jiaozhou Road, Shanghai, which caused a great damage to the building. 58 people died in this fire Considering the surviving methods are limited, we four came up with a new idea, parachute jumping. It means that in every house, some parachutes are needed to be equipped. When fire or earthquake takes place, this parachute jumping method might be of some help. However, this method is dangerous to some degree, because it takes time for a person to unfasten the parachute, if he unfastens it late, the landing velocity Source: http://www.doksinet will be a little big so that his body may not be able to endure the big impulsive force. According to this security problem, we have made some analysis to solve it. II. Analysis procedure 1. Part One We only consider one-dimension condition,

ignoring wind shear in the transverse dimension. Let’s consider such a problem. A person is jumping from a high building of which height is H. During the distance of x0, he is opening the parachute and falling as a free-falling body. When the falling distance reaches x0, the parachute is opened (we call it “opening point”). The air resistance is R, and its coefficient is k All the variables are listed below: Variables: Falling distance：x x0 x Velocity：v Distance to open the parachute：x0 H Velocity when opening the parachute：v0 Total height of the building：H Time：t Acceleration：a Total mass：m Acceleration of gravity：g Coefficient of air resistance：k Air resistance：R 1) Equation of motion before the opening point (Free-falling body, ignoring the air resistance)： 2) Assume the parachute is opened at the distance of x0, then the velocity when opening the parachute satisfies this equation： 3) Air resistance equation： K is the coefficient of air

resistance. The value of k can be obtained in this way: Let the maximum velocity of the falling body in the condition of gravity in the air equal to the maximum safety speed which can ensure the jumper to land safely. Moreover, we Source: http://www.doksinet assume the total mass of the jumper and the equipment is 80+7=87KG( Taking person with a litter larger weight into account, we assume his mass is 80KG). After looking up reference materials and information, we can get the maximum safety speed is 8.5m/s, namely: So k=11.8kg/s 4) Force equilibrium equation of falling after the opening point： Initial condition：v= v0 x= x0 Using separation variables method to solve this ODE, we obtain the relationship between v and x： 5) Discussion of sign of the total acceleration ： It takes at least 2 seconds from the moment the man begins to jump to the parachute is totally opened, and the velocity of the free-falling body after falling 2 seconds is 2*9.8=196m/s>85s So must be

negative，which means that after the parachute is opened, the jumper must come through a deceleration course. 6 ) Integral the equation after separation variables: C is determined by x0 and v0： Take C into the equation： Till now, this ODE has been solved. 2. Part two According to high building parachute jumping, total falling distance is the H, and Source: http://www.doksinet the maximum velocity of landing is 8.5m/s If we take H and v=85m/s into the equation solved above, we can get the corresponding x0, which is the maximum safety distance of opening parachute. Change the solved equation in part one: According the condition of the existing high building in China, we can obtain different x0 corresponding to different height of the building (Assume the jumper begins to jump at the top of the building). Those buildings are: Shanghai Jin Mao Tower, Shanghai World Financial Center, Taipei 101 Building, CCTV Building and teacher’s apartment on Jiao Zhou Road. Use Matlab to

get the solution: x=solve(-87/2/11.8+x+87/2/118*ln(11.8/87*29.8*x-9.8)=3401,x) 1) Shanghai Jin Mao Tower 88th floor sightseeing flat: Height is 340.1m x0=318.97m 2) Shanghai World Financial Center 100th floor sightseeing flat: Height is 477.96m x0=455.51m 3) Taipei 101 Building： Height is 439.2m x0=417.07 4) CCTV Building Height is 234m x0=214.36 5) Teacher’s apartment on Jiao Zhou Road. Height is 85m x0=69.64 3. Part three It takes some time for the jumper to adjust himself and open the parachute, and we assume this minimum free-falling period of time is tmin. So the velocity when the parachute is opened is v0=gtmin, and falling distance is x0=0.5*gt2min m. As the landing speed is the maximum safety speed v=8.5m/s，we can obtain the minimum Hmin referring to the tmin. 1、 tmin=3s 2、 tmin=4s 3、tmin=5s t=3s t=4s Hmin =h=57.65m Hmin =h=94.22m Source: http://www.doksinet t=5s Hmin =h=140.03m III．Discussion and conclusion In our analysis, we considered two conditions.

One is that the total falling distance H is known, and we can get the free-falling distance x0, which is the maximum safety distance of opening parachute. The other is that the minimum free-falling period of time is tmin is known, and we can get the minimum total falling distance Hmin, which helps the architect to set a proper jumping flat with proper height h according to Hmin. Of course, some improvements are still in need in our project. Firstly, our project is an ideal one-dimensional model which ignores the shear force of wind. However, this kind of force can’t be ignored in the certain height. Secondly, for different person, the opening time may be of some difference. Our project should cover a wide range of data for different people. Thirdly, a new kind of parachute is necessary to satisfy the need of high building escaping. It should be easy to use for everyone and can open in a short time. Nowadays, the escaping and rescuing of high building fire disaster is still one of the

greatest problems in the world. So we set up our mind to find an efficient and economic way to solve this problem. We hope our method can be useful someday in the future